1 / 3

Lab 1. 级数计算 [Hamming (1962)] x = 0.0, 0.1, 0. 5 ,1.0; , 10.0, 10 0.0, 2 00.00. 绝对误差小于 1.0e-6.

Lab 1. 级数计算 [Hamming (1962)] x = 0.0, 0.1, 0. 5 ,1.0; , 10.0, 10 0.0, 2 00.00. 绝对误差小于 1.0e-6. 输出  两列输出： x 和  ( x ) 如 C fprintf: fprintf(outfile,“%6.2f  ,  %16.12 e <br>&quot;,x,psix); /* here  represents a space */. Sample Output (  represents a space)

kisha
Download Presentation

Lab 1. 级数计算 [Hamming (1962)] x = 0.0, 0.1, 0. 5 ,1.0; , 10.0, 10 0.0, 2 00.00. 绝对误差小于 1.0e-6.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lab 1. 级数计算[Hamming (1962)] x = 0.0,0.1,0.5,1.0; ,10.0,100.0,200.00.绝对误差小于1.0e-6. 输出  两列输出： x和 (x) 如 C fprintf: fprintf(outfile,“%6.2f,%16.12e\n",x,psix);/* hererepresents a space */

  2. Sample Output ( represents a space) 0.00,0.644934066848e+001 0.10,0.534607244904e+001 ... 1.00,0.99999900000e+001 ...

  3. 提示： 例如，当 时，该级数和可手工求出，据此可估算出前n项部分和的绝对误差。 此外，不难看出当 值越大时， 其前n项部分和的绝对误差是越来越小的。

More Related